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Anna [14]
3 years ago
11

Unpolarized light with intensity I0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polariz

ing filter whose axis is at 42.0 ∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer.
Physics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

0.276I_0

Explanation:

When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.

So, after the first polarizer, the intensity of light passing through is:

I_1=\frac{I_0}{2}

Where I_0 is the initial intensity of the unpolarized light.

Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:

I_2=I_1 cos^2 \theta

where:

\theta is the angle between the axes of the two polarizers

In this problem the angle is

\theta=42^{\circ}

So the intensity after of light the 2nd polarizer is:

I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0

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A heavier car is always safer in a crash than a lighter car.
kvv77 [185]

Answer:

not true because the mass from the heavy car will cause it to damage more

Explanation:

4 0
3 years ago
A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

8 0
3 years ago
I really need help on this question!​
steposvetlana [31]

Answer:

option "c"

Explanation:

because in gases molecules are further apart and move very quickly

4 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi
mr_godi [17]

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of  artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

8 0
3 years ago
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